Simplify; express your answer in exponential form. Assume $r\neq 0, a\neq 0$. $\dfrac{{(r^{5}a^{5})^{-1}}}{{(r^{-5}a^{-1})^{-5}}}$
Explanation: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(r^{5}a^{5})^{-1} = (r^{5})^{-1}(a^{5})^{-1}}$ On the left, we have ${r^{5}}$ to the exponent ${-1}$ . Now ${5 \times -1 = -5}$ , so ${(r^{5})^{-1} = r^{-5}}$ Apply the ideas above to simplify the equation. $\dfrac{{(r^{5}a^{5})^{-1}}}{{(r^{-5}a^{-1})^{-5}}} = \dfrac{{r^{-5}a^{-5}}}{{r^{25}a^{5}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-5}a^{-5}}}{{r^{25}a^{5}}} = \dfrac{{r^{-5}}}{{r^{25}}} \cdot \dfrac{{a^{-5}}}{{a^{5}}} = r^{{-5} - {25}} \cdot a^{{-5} - {5}} = r^{-30}a^{-10}$